Question

How do you multiply $\left( 8x-5 \right)\left( 4x-7 \right)$ ?

Answer 1

Hint: We can multiply both the term given in the question by simple algebraic multiplication we know the $a\left( b+c \right)=ab+ac$ and property of exponential function ${{x}^{n}}{{x}^{m}}={{x}^{m+n}}$ , so the multiplication of x with x gives us ${{x}^{2}}$ . Using above both formulas we can multiply$\left( 8x-5 \right)\left( 4x-7 \right)$.

Complete step by step answer:
We have multiply $\left( 4x-7 \right)$ with $\left( 8x-5 \right)$ we know the algebraic formula $a\left( b+c \right)=ab+ac$
So assume a as $\left( 8x-5 \right)$ , b as 4x and c as -7 , we can apply the above algebraic formula
 $\left( 8x-5 \right)\left( 4x-7 \right)=4x\left( 8x-5 \right)-7\left( 8x-5 \right)$
Now in first term in RHS $4x\left( 8x-5 \right)$ and in the second term in RHS $-7\left( 8x-5 \right)$we can again apply the formula
So applying the formula we get
 $\Rightarrow \left( 8x-5 \right)\left( 4x-7 \right)=32{{x}^{2}}-20x-56x+35$
Adding -20x and -56x we get -76x
$\Rightarrow \left( 8x-5 \right)\left( 4x-7 \right)=32{{x}^{2}}-76x+35$
So the multiplying $\left( 4x-7 \right)$ with $\left( 8x-5 \right)$ we get $32{{x}^{2}}-76x+35$
We can say the factor form of $32{{x}^{2}}-76x+35$ is $\left( 8x-5 \right)\left( 4x-7 \right)$

Note: We have seen how we multiplied 2 linear terms, if the terms are not linear then also we will apply the same process for the multiplication, just remember the exponential formula ${{x}^{n}}{{x}^{m}}={{x}^{m+n}}$ . If there are more than 2 terms instead of 2 terms then we can multiply first 2 terms then multiply the result with the next term and this goes on until the last term.
We know the algebraic formula $a\left( b+c \right)=ab+ac$ , keep in mind that if we reverse the plus and multiplication sign then the formula will be invalid, that means $a+\left( b\times c \right)\ne \left( a+b \right)\times \left( a+c \right)$
In the mutilation of the algebraic terms the coefficient of x in the result is equal to the sum of coefficient of x in all the terms.